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3 years ago

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a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

moves right at 0.800 m/s. find the momentum of the 155 kg car afterwards. PLEASE HELP

1 answer:

Vaselesa [24]3 years ago

7 0

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

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The speed of an object in a mass-spring system is given under the function

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3 years ago

A ball is thrown straight up into the air with an initial velocity of 50 ft/sec. The height h(t) of the ball after t seconds is

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Answer:

The average velocity for the time period beginning when t=1 and lasting 0.1 seconds = 16.40 ft/s.

Explanation:

Given that the height of the ball at time t is

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The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

If $\rm h_1,\ h_2$ are the heights of the ball at time $\rm t_1$ and $\rm t_2$ , then the total displacement covered by the ball from time $\rm t_2$ to $\rm t_1$ is $\rm h_2-h_1$ .

Thus, the average velocity of the ball for the time interval $\rm t_2-t_1$ is given by

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$\rm t_1=1\ sec.\\t_2 = 1\sec + 0.1\ sec = 1.1\ sec.\\\\h_1=50\times 1-16\times 1^2=34\ ft.\\h_2=50\times 1.1-16\times 1.1^2=35.64\\\\Therefore,\\\\v_{av} = \dfrac{35.64-34}{1.1-1}=16.40\ ft/s.$

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.30 is the least precise measurement so round to two decimal points

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3 years ago

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Answer:

<em>a) Jack does more work uphill</em>

<em>b) Numerically, we can see that Jill applied the most power downhill</em>

<em></em>

Explanation:

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Jill's mass = $1.5x = 75$

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a) work done by Jack uphill = mgh

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<em>this shows that Jack does more work climbing up the hill</em>

<em></em>

b) assuming Jack's time downhill to be t,

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P = $\frac{work}{time}$

For Jack, power = $\frac{11036.25}{t}$

For Jill, power = $\frac{3*7357.5}{t}$ = $\frac{22072.5}{t}$

<em>Numerically, we can see that Jill applied the most power downhill</em>

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